Question: Estimating $e^{-0.95}$ using a Maclaurin polynomial, what is the least degree of the polynomial that assures an error smaller than $0.001$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $4$ (Choice C) C $5$ (Choice D) D $6$
We will use the Lagrange error bound. Let's assume the polynomial's degree is $n$. The $(n+1)^{\text{th}}$ derivative of $e^x$ is $e^x$. On the interval between $x=-0.95$ and $x=0$, the greatest value of the derivative is $e^{0}=1$. The Lagrange bound for the error assures that: $|R_n(-0.95)|\leq \left| \dfrac{1}{(n+1)!}(-0.95)^{n+1} \right|$ Solving $\dfrac{1}{(n+1)!}0.95^{n+1}<0.001$ using trial and error, we find that $n\geq 6$. In conclusion, the least degree of the polynomial that assures our error bound is $6$.